IndexError字符串索引超出范围

Question

s="(8+(2+4))"
def checker(n):
if len(n) == 0:
    return True
if n[0].isdigit==True:
    if n[1].isdigit==True:
        return False
    else:
        checker(n[1:])
else:
    checker(n[1:])

This is what I have so far. Simple code, trying to see if a string meets the following conditions. However when i perform checker(s) i get:

True
IndexError: string index out of range

Any help? Thanks in advance Edit: The function's purpose is to produce true if the string contains only single digit numbers, and false if 2 or more-figured digits exist in the string.

Answer 1

When the length of n is 0, the n[0] part is going to raise an error because the string in empty. You should add a return statement there instead of print.

def checker(n):
    if len(n) < 2:
        return True
    if n[0] in x:

Note that the conditions must be len(n) < 2 otherwise you'll get an error on n[1] when the length of string is 1.

Secondly you're trying to match characters to a list which contains integers, so the in checks are always going to be False . Either convert the list items to string or better use str.isdigit .

>>> '1'.isdigit()
True
>>> ')'.isdigit()
False
>>> '12'.isdigit()
True

---

Update:

You can use regex and all for this:

>>> import re
def check(strs):
    nums = re.findall(r'\d+',strs)
    return all(len(c) == 1 for c in nums)
... 
>>> s="(8+(2+4))"
>>> check(s)
True
>>> check("(8+(2+42))")
False

---

Working version of your code:

s="(8+(2+4))"
def checker(n):
    if not n:           #better than len(n) == 0, empty string returns False in python
        return True
    if n[0].isdigit():  #str.digit is a method and it already returns a boolean value   
        if n[1].isdigit():   
            return False
        else:
            return checker(n[1:])  # use return statement for recursive calls
                                   # otherwise the recursive calls may return None  
    else:
        return checker(n[1:])        

print checker("(8+(2+4))")
print checker("(8+(2+42))")

output:

True
False

Answer 2

You should do return True after the first if statement, not print True . The function continues to run after that statement and hits an error when the input is size 0.

Answer 3

I can't reproduce your error.

I had to fix a few things:

• Indentation, which I'm guessing was just a problem pasting into the page
• .isdigit() is a function; calling .isdigit==True is going to compare a function object with True, which is never going to be true. I changed .isdigit==True to .isdigit()
• I made sure return value gets bubbled up - without this, the recursion completes okay but the outermost function just returns "None".

Aside from that, a few print statements show that this is working as expected

s="(8+(2+4))"
t="(8+(20+4))"
def checker(n):
  print "checking %s" % n
  if len(n) == 0:
    print "Returning true"
    return True
  if n[0].isdigit():
    if n[1].isdigit():
        print "returning false"
        return False
    else:
        return checker(n[1:])
  else:
    return checker(n[1:])

print checker(s)
print checker(t)

Output:

checking (8+(2+4))
checking 8+(2+4))
checking +(2+4))
checking (2+4))
checking 2+4))
checking +4))
checking 4))
checking ))
checking )
checking 
Returning true
True
checking (8+(20+4))
checking 8+(20+4))
checking +(20+4))
checking (20+4))
checking 20+4))
returning false
False